AS June 2023 Q1
1.
| Scheme | Marks | AO |
|---|---|---|
| E.g. \((x \neq 2 \Rightarrow)\,5x = 12(x - 2) \Rightarrow x = \ldots\) \(\dfrac{5x}{x - 2} \geqslant 12 \Rightarrow 5x(x - 2) \geqslant 12(x - 2)^2 \Rightarrow (x - 2)(5x - 12(x - 2)) \geqslant 0 \Rightarrow x = \ldots\) or \(\dfrac{5x}{x - 2} \geqslant 12 \Rightarrow \dfrac{5x - 12(x - 2)}{x - 2} \geqslant 0 \Rightarrow x = \ldots\) o.e. | M1 | 2.1 |
| E.g. \(x = 2\) and \((x \neq 2 \Rightarrow)\,5x = 12(x - 2) \Rightarrow x = \ldots\) \((x - 2)(-7x + 24) \geqslant 0\) or \((x - 2)(7x - 24) \leqslant 0 \Rightarrow x = \ldots, \ldots\) or \(7x^2 - 38x + 48 \leqslant 0\) or \(-7x^2 + 38x - 48 \geqslant 0 \Rightarrow x = \ldots, \ldots\) or \(\dfrac{24 - 7x}{x - 2} \geqslant 0 \Rightarrow x = \ldots, \ldots\) | dM1 | 1.1b |
| Critical values \(x = 2, \dfrac{24}{7}\) | A1 | 1.1b |
| \(2 \lt x \leqslant \dfrac{24}{7}\) | A1 | 2.3 |
| (4) |
Notes
M1: For an algebraic method to find the critical value aside from \(x = 2\). May set equal and use \(x \neq 2\) or \(x \lt 2\) and \(x \gt 2\) to find the \(x\) coordinate of the intersection of line and curve, or may multiply through by \((x - 2)^2\) and gather terms onto one side and solve the quadratic, or gather all terms onto one side and put over a common denominator and solve the quadratic. Allow with any inequality or equality for the first two marks.
dM1: Finds both the critical values by a correct algebraic method (allowing for slips rearranging). The \(x = 2\) may just be stated or used as a boundary value of the interval. Use of quadratic formula - usual rules. Dependent on previous method mark.
A1: Correct critical values stated or used in solution. Allow awrt 3.43 for \(\dfrac{24}{7}\) for this mark. A0 if other values also used.
A1: Deduces the correct inequality. Must be exact. Accept alternative notations, such as \(\left(2, \dfrac{24}{7}\right]\) but formal set notation is not required - score the inequality given. Accept as separate inequalities.
If \(2 \lt x \leqslant \dfrac{24}{7}\) follows \(2 \leqslant x \leqslant \dfrac{24}{7}\) arising from \((x - 2)(7x - 24) \leqslant 0\) (oe) then allow A1 b.o.d. that the latter answer is a rejection of the 2 being included.
(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)
| Scheme | Marks | AO |
|---|---|---|
| \(x = 3\) | B1 | 2.2a |
| (1) | ||
| (5 marks) |
Notes
B1cao: Deduces the correct value for \(x\) and no other values as long as 3 is in their solution set from (a). Allow if the endpoint 2 was included in their answer to (a) as long as it is not given as a solution for (b).